hdu2588——GCD(欧拉函数)

GCD

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2136    Accepted Submission(s): 1087

Problem Description

The greatest common divisor GCD(a,b) of two positive integers a and b,sometimes written (a,b),is the largest divisor common to a and b,For example,(1,2)=1,(12,18)=6.
(a,b) can be easily found by the Euclidean algorithm. Now Carp is considering a little more difficult problem:
Given integers N and M, how many integer X satisfies 1<=X<=N and (X,N)>=M.

 

Input

The first line of input is an integer T(T<=100) representing the number of test cases. The following T lines each contains two numbers N and M (2<=N<=1000000000, 1<=M<=N), representing a test case.

 

Output

For each test case,output the answer on a single line.

 

Sample Input

31 110 210000 72

 

Sample Output

16260

 

Source

ECJTU 2009 Spring Contest 

题意大概：给定N,M(2<=N<=1000000000, 1<=M<=N), 求1<=X<=N 且gcd(X,N)>=M的个数。

解法：数据量太大，用常规方法做是行不通的。后来看了别人的解题报告说，先找出N的约数x，

          并且gcd（x，N）>= M，结果为所有N/x的欧拉函数之和。

          因为x是Ｎ的约数，所以ｇｃｄ（ｘ，Ｎ）＝ｘ >= M；

　　　设ｙ＝Ｎ／ｘ，ｙ的欧拉函数为小于ｙ且与ｙ互质的数的个数。

　　　设与ｙ互质的的数为ｐ１，ｐ２，ｐ３，…，ｐ４

　　　那么ｇｃｄ（ｘ* pi，N）= x >= M。

          也就是说只要找出所有符合要求的y的欧拉函数之和就是答案了。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
int oula(int n)
{
    int res=1,i;
    for(i=2; i*i<=n; i++)
        if(n%i==0)//i是素数
        {
            n/=i;//因为要算ni-1次方，所以在这先除一下
            res*=i-1;
            while(n%i==0)
            {
                n/=i;//n越来越小
                res*=i;
            }
        }
    if(n>1)
        res*=n-1;
    return res;
}
int main()
{
    int i,j,k,t;
    int n,m;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&m);
        int ans=0;
        for(i=1;i*i<=n;i++){
            if(n%i==0){
                if(i>=m)ans+=oula(n/i);
                if(n/i>=m&&i!=n/i)ans+=oula(n/(n/i));//注意平方根不能重复算
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}